1986 AIME Problem 5

Attempt Problem 5 of the 1986 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1986 AIME solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

What is the largest positive integer

n

for which

n^3+100

is divisible by

n+10?

Answer

Reduce

n^3+100

modulo

n+10

The condition makes

n+10

a positive divisor of a fixed integer

Modulo

n+10,

we have

n\equiv-10.

Thus

\begin{gathered} n^3+100 \equiv(-10)^3+100\\ \equiv-900\pmod{n+10}. \end{gathered}

The condition is therefore equivalent to

900

being divisible by

n+10.

Since

n

is positive,

n+10

is a positive divisor of

900,

and its largest possible value is

900.

This gives

n=900-10=890.

Problem 5 in other years